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Violation of Newton's Third Law

  1. Oct 14, 2005 #1
    Are there any examples of violation of Newton's third law?
    The two examples that I know are :
    1) Two masses moving relativistically apart from each other.
    2) Two charges moving in mutulally perpendicular directions.
    Can anybody further elaborate on these exmaples with proper explanation and any other examples in which the third law is violated.
     
  2. jcsd
  3. Oct 14, 2005 #2

    ZapperZ

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    Sorry? How are these two examples "violate" Newton's 3rd Law? And why just that law, considering all 3 Newton's laws are one of the SAME?

    Zz.
     
  4. Oct 14, 2005 #3

    ahrkron

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    I think the second example is mentioned in Goldstein's classical mechanics textbook. When two charges are in motion so as to "cross a T", they do not experience "equal and opposite" forces from their interaction. This is a result of the fact that EM forces do not generally point in the direction of the line that joins the particles in question.
     
  5. Oct 14, 2005 #4

    ZapperZ

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    But isn't this more to do with the retardation effects than "violation of the 3rd law"?

    Zz.
     
  6. Oct 14, 2005 #5

    jtbell

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    That's because the electromagnetic field carries both momentum and energy.
     
  7. Oct 14, 2005 #6
    Violation of Newton's Third Law.

    The examples that I have quoted are from the book Elementary Physics by I. P. Guruskii (Mir Publishers 1987) pg. 55.

    How do these examples violate the law?

    That is what I want to know ...

    The book does not give any explanation for these 'violations', nor does it give any further examples. Does any body know about a reference which discusses the 'violations' of the law in much details.
     
  8. Oct 14, 2005 #7
    As we all know Newton's Third Law is:

    Forces occur in equal and opposite pairs: whenever
    object A exerts a force on object B, object B must also
    be exerting a force on object A. The two forces are
    equal in magnitude and opposite in direction.
    This can be shown as;

    F(A on B) = –F (B on A)

    So far, no violation of the third law has ever
    been discovered, whereas the first and second laws were shown to have
    limitations by Einstein and planck etc... but there is definatly no record of any violation of the 3rd law

    Hopefully this will put your mind at rest:smile: :smile:
     
  9. Oct 16, 2005 #8

    Stingray

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    Newton's 3rd law is necessarily violated by any relativistic theory (except in special cases). Standard Maxwell electrodynamics and general relativity are examples. I suppose you could wiggle out of it by making contorted definitions, but the usual ones show very easily that Newton's 3rd law cannot be exact.

    Think about two objects separated by a considerable distance. At any instant of time, the impossibility of faster than light signalling requires that particle A cannot know what particle B is doing at that time. There is actually a finite time interval during which A is completely ignorant of B, and vice versa. During this window, either particle is free to do whatever it wants (within the laws of physics) without immediately affecting the other one. This freedom of motion translates into a (usually small) freedom in the forces felt at the particular instant of time we are considering. This implies violations of the 3rd law.

    To be more specific, let A and B be initially fixed a distance d apart. At a time less than d/c in past, A starts moving. Since the field sourced by B is not homogeneous, the force exerted on A depends on exactly how it moves. But the field sourced by A around B has not (yet) changed, so the force felt by B is independent of A's motion (for a short time). The force on A therefore varies while the force on B does not. The two cannot be always be equal and opposite.

    By the way, this violation is exactly what is needed for bodies to exert net forces and torques on themselves. This most often arises in discussions of radiation reaction effects, although it usually isn't introduced in exactly this way. The math gets ugly very quickly, but it does work out to be the same as derivations using energy-momentum conservation.
     
  10. Oct 16, 2005 #9

    HallsofIvy

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    "The field sourced by A"? You mean like the gravitational field or the electromagnetic field? What you have shown is that there cannot be an "instantaneous" reaction at B to a change in position of A. It is certainly well know that a change in the position of a charged partical at A, causing a change in the electromagnetic field, does not effect B until a certain time has elapsed- the distance between A and B divided by the speed of light because waves in the electromagnetic field ARE light. It is firmly believed, but, I think, not yet supported by direct experimental evidence that changes in the gravitational field are also transmitted at the speed of light. Pretty much what your argument implies. It certainly does NOT contradict Newton's third law.
     
  11. Oct 16, 2005 #10

    Stingray

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    Yes.

    Yes.

    Huh? Yes it does. My example showed that the force exerted "by B on A" at any fixed time can be varied without affecting the force "by A on B." The two obviously can't be equal and opposite if it's impossible for them to even be perfectly correlated. If you say that doesn't violate Newton's 3rd law, I'd like to know what version of it you're thinking of.
     
  12. Oct 16, 2005 #11

    Stingray

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    Also, this hardly takes sophisticated arguments. Every E&M book gives an example of a situation where forces between two particles are not equal and opposite. I don't understand the point of Zz's dismissal that it's from "retardation effects." So what? That's an important part of Maxwell's theory. The 3rd law states that the forces should be equal and opposite, but they're not. So it's wrong. Of course it's a pretty good approximation in most situations, but that's not what we're talking about.
     
  13. Oct 16, 2005 #12

    ZapperZ

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    Er.... So what?

    What is applied via the retarded effects are NOT within the realm of what is described via the three Newtonian Laws. If we extend it further, we can find MORE things that are not accurately described by these laws. So why pick on the 3rd Law in particular?

    Newtonian laws assume instantaneous, action-at-a-distant effects. They do not take into account retarded effects. Elementary E&M don't either until you start considering the dynamics of the field equations! So to say that the 3rd Law fails when it is applied to a situation it wasn't MEANT to be applied is puzzling. That's like accusing that Thermodynamics 2nd Law is being violated under certian non-equilibrium conditions. What gives? It wasn't meant to be applied there in the first place!

    Zz.
     
  14. Oct 16, 2005 #13

    Stingray

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    Ok, I understand you now. Our disagreement was just semantics.
     
  15. Oct 16, 2005 #14

    pervect

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    Here's a applicable quote from Goldstein's "Classical mechanics", pg 7.

     
  16. Oct 17, 2005 #15

    vanesch

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    In a way, we can equate Newton's 3rd law with "conservation of momentum". Indeed, if the "force on A" is dp1/dt and the "force on B" is dp2/dt and one says that these forces must be equal and opposite, this means that:
    d(p1+p2)/dt = 0, so p1+p2 is constant.
    The examples of relativistic charges not exerting "equal and opposite" forces on eachother is of course correct, but it is not because "Newton's third law" is failing, it is because we've missed a player in the field (:smile:): the EM field itself.
    So we naively think that particle A is exerting a force on particle B (Newtonian action at a distance), but we're missing that particle A is in fact NOT interacting with particle B, but only with the local EM field, and that this local EM field has its own dynamics, which results in the end in it interacting with particle B.
    So this is in fact equivalent with a totally "Newtonian" picture where particle A hits particle C, which flies with finite velocity towards particle B.
    If you forget to include particle C, you will find a "violation of Newton's third law" when particle C is travelling.
    cheers,
    Patrick.
     
  17. Oct 17, 2005 #16

    pervect

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    Going back to the original question, if you take a pair of charges at a 45 degree angle, and do a "Loerntz boost", the force between the charges initally obeys both the strong and weak forms of the third law (as per Goldstein's terminology), as the forces are equal, opposite, and on a direct line between the charges.

    If you do a "Lorentz boost" to a new frame, the force between the charges satisfies the weak law, but not the strong law. This means that the forces are equal and opposite, but no longer on a direct line between the charges in the frame. (Note that we are talking about the 3-forces here!).

    See the attached diagram for how it all turns out.

    This gives rise to a rather obscure paradox in special relativity known as the "lever paradox", whereby internal forces contribute to the angular momentum of a system.

    http://scitation.aip.org/getabs/ser...00043000007000615000001&idtype=cvips&gifs=yes
     

    Attached Files:

  18. Aug 9, 2010 #17
    The quote from Goldstein applies to angular momentum - not to linear momentum. And sure enough, in the usual example of two charged particles moving apart at right angles angular momentum is conserved. But maybe linear not? The only escape there is the A.q factor in the momentum of a charged particle. Works in some cases - but not all :wink:. On average in a plasma there will be momentum conserved and energy at any rate, though.
     
  19. Dec 26, 2010 #18
    Funny: all those examples in Goldstein, Feynmann, Griffith et al. consider only the obvious case of charged particles in flight. A current in a wire is not like that at all as the EM field cannot be wheeled out as an excuse for non-violation of Newton's Third Law. I.e. the wire is electrically neutral as the same number of electrons enter it as leave per second, and they are balanced by the positive metal ions. So no radial E field. Only E field is along the wire. So the EM field, if it exists at all, would be radial, which does not balance the forces in the planes of the crossing wires.
     
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